How do you solve Jump Game II on LeetCode?

Jump Game II (LeetCode #45) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The greedy approach significantly reduces the number of computations by focusing on the farthest reachable index.

What is the brute force solution for Jump Game II?

The brute force approach for Jump Game II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible jump from each index to find the minimum number of jumps to reach the last index. This method is straightforward but inefficient as it explores all paths. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Jump Game II?

Jump Game II uses the following concepts: Array, Dynamic Programming, Greedy. The recommended patterns to study are: Greedy, Dynamic Programming, Array.

What are the interview tips for Jump Game II?

Always clarify the problem constraints and edge cases before diving into coding. Think about how you can optimize a brute-force solution by reducing unnecessary calculations. Practice explaining your thought process as you code, as this helps interviewers understand your reasoning.

What are common mistakes when solving Jump Game II?

Not considering the case where you can jump directly to the last index from an earlier index. Overcomplicating the problem by trying to track every possible path instead of focusing on the farthest reachable index.

#45

Jump Game II

Medium

ArrayDynamic ProgrammingGreedyGreedyDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution uses a greedy approach to minimize the number of jumps. Instead of exploring all paths, it keeps track of the farthest point reachable with the current number of jumps and updates the jump count accordingly.

⚙️

Algorithm

3 steps

1Step 1: Initialize variables for the current end of the jump range and the farthest reachable index.
2Step 2: Iterate through the array, updating the farthest index reachable from the current position.
3Step 3: When reaching the end of the current jump range, increment the jump count and update the current end to the farthest index.

solution.py10 lines

1def jump(nums):
2    jumps = 0
3    current_end = 0
4    farthest = 0
5    for i in range(len(nums) - 1):
6        farthest = max(farthest, i + nums[i])
7        if i == current_end:
8            jumps += 1
9            current_end = farthest
10    return jumps

ℹ

Complexity note: This complexity is linear because we only make a single pass through the array, updating our jump count and farthest index without any nested loops.

1The greedy approach significantly reduces the number of computations by focusing on the farthest reachable index.
2Understanding the problem as a series of intervals can help visualize the jumps needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.